3.13.63 \(\int \frac {x^2}{(a-b x^4)^{5/4}} \, dx\) [1263]

Optimal. Leaf size=81 \[ \frac {1}{b x \sqrt [4]{a-b x^4}}-\frac {\sqrt [4]{1-\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt {b} \sqrt [4]{a-b x^4}} \]

[Out]

1/b/x/(-b*x^4+a)^(1/4)-(1-a/b/x^4)^(1/4)*x*(cos(1/2*arccsc(x^2*b^(1/2)/a^(1/2)))^2)^(1/2)/cos(1/2*arccsc(x^2*b
^(1/2)/a^(1/2)))*EllipticE(sin(1/2*arccsc(x^2*b^(1/2)/a^(1/2))),2^(1/2))/(-b*x^4+a)^(1/4)/a^(1/2)/b^(1/2)

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Rubi [A]
time = 0.03, antiderivative size = 81, normalized size of antiderivative = 1.00, number of steps used = 5, number of rules used = 5, integrand size = 16, \(\frac {\text {number of rules}}{\text {integrand size}}\) = 0.312, Rules used = {293, 319, 342, 281, 234} \begin {gather*} \frac {1}{b x \sqrt [4]{a-b x^4}}-\frac {x \sqrt [4]{1-\frac {a}{b x^4}} E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt {b} \sqrt [4]{a-b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Int[x^2/(a - b*x^4)^(5/4),x]

[Out]

1/(b*x*(a - b*x^4)^(1/4)) - ((1 - a/(b*x^4))^(1/4)*x*EllipticE[ArcCsc[(Sqrt[b]*x^2)/Sqrt[a]]/2, 2])/(Sqrt[a]*S
qrt[b]*(a - b*x^4)^(1/4))

Rule 234

Int[((a_) + (b_.)*(x_)^2)^(-1/4), x_Symbol] :> Simp[(2/(a^(1/4)*Rt[-b/a, 2]))*EllipticE[(1/2)*ArcSin[Rt[-b/a,
2]*x], 2], x] /; FreeQ[{a, b}, x] && GtQ[a, 0] && NegQ[b/a]

Rule 281

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> With[{k = GCD[m + 1, n]}, Dist[1/k, Subst[Int[x^((m
 + 1)/k - 1)*(a + b*x^(n/k))^p, x], x, x^k], x] /; k != 1] /; FreeQ[{a, b, p}, x] && IGtQ[n, 0] && IntegerQ[m]

Rule 293

Int[(x_)^2/((a_) + (b_.)*(x_)^4)^(5/4), x_Symbol] :> -Simp[(b*x*(a + b*x^4)^(1/4))^(-1), x] - Dist[1/b, Int[1/
(x^2*(a + b*x^4)^(1/4)), x], x] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 319

Int[1/((x_)^2*((a_) + (b_.)*(x_)^4)^(1/4)), x_Symbol] :> Dist[x*((1 + a/(b*x^4))^(1/4)/(a + b*x^4)^(1/4)), Int
[1/(x^3*(1 + a/(b*x^4))^(1/4)), x], x] /; FreeQ[{a, b}, x] && NegQ[b/a]

Rule 342

Int[(x_)^(m_.)*((a_) + (b_.)*(x_)^(n_))^(p_), x_Symbol] :> -Subst[Int[(a + b/x^n)^p/x^(m + 2), x], x, 1/x] /;
FreeQ[{a, b, p}, x] && ILtQ[n, 0] && IntegerQ[m]

Rubi steps

\begin {align*} \int \frac {x^2}{\left (a-b x^4\right )^{5/4}} \, dx &=\frac {1}{b x \sqrt [4]{a-b x^4}}+\frac {\int \frac {1}{x^2 \sqrt [4]{a-b x^4}} \, dx}{b}\\ &=\frac {1}{b x \sqrt [4]{a-b x^4}}+\frac {\left (\sqrt [4]{1-\frac {a}{b x^4}} x\right ) \int \frac {1}{\sqrt [4]{1-\frac {a}{b x^4}} x^3} \, dx}{b \sqrt [4]{a-b x^4}}\\ &=\frac {1}{b x \sqrt [4]{a-b x^4}}-\frac {\left (\sqrt [4]{1-\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {x}{\sqrt [4]{1-\frac {a x^4}{b}}} \, dx,x,\frac {1}{x}\right )}{b \sqrt [4]{a-b x^4}}\\ &=\frac {1}{b x \sqrt [4]{a-b x^4}}-\frac {\left (\sqrt [4]{1-\frac {a}{b x^4}} x\right ) \text {Subst}\left (\int \frac {1}{\sqrt [4]{1-\frac {a x^2}{b}}} \, dx,x,\frac {1}{x^2}\right )}{2 b \sqrt [4]{a-b x^4}}\\ &=\frac {1}{b x \sqrt [4]{a-b x^4}}-\frac {\sqrt [4]{1-\frac {a}{b x^4}} x E\left (\left .\frac {1}{2} \csc ^{-1}\left (\frac {\sqrt {b} x^2}{\sqrt {a}}\right )\right |2\right )}{\sqrt {a} \sqrt {b} \sqrt [4]{a-b x^4}}\\ \end {align*}

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Mathematica [C] Result contains higher order function than in optimal. Order 5 vs. order 4 in optimal.
time = 7.72, size = 55, normalized size = 0.68 \begin {gather*} \frac {x^3 \sqrt [4]{1-\frac {b x^4}{a}} \, _2F_1\left (\frac {3}{4},\frac {5}{4};\frac {7}{4};\frac {b x^4}{a}\right )}{3 a \sqrt [4]{a-b x^4}} \end {gather*}

Antiderivative was successfully verified.

[In]

Integrate[x^2/(a - b*x^4)^(5/4),x]

[Out]

(x^3*(1 - (b*x^4)/a)^(1/4)*Hypergeometric2F1[3/4, 5/4, 7/4, (b*x^4)/a])/(3*a*(a - b*x^4)^(1/4))

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Maple [F]
time = 0.02, size = 0, normalized size = 0.00 \[\int \frac {x^{2}}{\left (-b \,x^{4}+a \right )^{\frac {5}{4}}}\, dx\]

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(-b*x^4+a)^(5/4),x)

[Out]

int(x^2/(-b*x^4+a)^(5/4),x)

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Maxima [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {Failed to integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-b*x^4+a)^(5/4),x, algorithm="maxima")

[Out]

integrate(x^2/(-b*x^4 + a)^(5/4), x)

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Fricas [F]
time = 0.07, size = 36, normalized size = 0.44 \begin {gather*} {\rm integral}\left (\frac {{\left (-b x^{4} + a\right )}^{\frac {3}{4}} x^{2}}{b^{2} x^{8} - 2 \, a b x^{4} + a^{2}}, x\right ) \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-b*x^4+a)^(5/4),x, algorithm="fricas")

[Out]

integral((-b*x^4 + a)^(3/4)*x^2/(b^2*x^8 - 2*a*b*x^4 + a^2), x)

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Sympy [C] Result contains complex when optimal does not.
time = 0.43, size = 39, normalized size = 0.48 \begin {gather*} \frac {x^{3} \Gamma \left (\frac {3}{4}\right ) {{}_{2}F_{1}\left (\begin {matrix} \frac {3}{4}, \frac {5}{4} \\ \frac {7}{4} \end {matrix}\middle | {\frac {b x^{4} e^{2 i \pi }}{a}} \right )}}{4 a^{\frac {5}{4}} \Gamma \left (\frac {7}{4}\right )} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x**2/(-b*x**4+a)**(5/4),x)

[Out]

x**3*gamma(3/4)*hyper((3/4, 5/4), (7/4,), b*x**4*exp_polar(2*I*pi)/a)/(4*a**(5/4)*gamma(7/4))

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Giac [F]
time = 0.00, size = 0, normalized size = 0.00 \begin {gather*} \text {could not integrate} \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

integrate(x^2/(-b*x^4+a)^(5/4),x, algorithm="giac")

[Out]

integrate(x^2/(-b*x^4 + a)^(5/4), x)

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Mupad [F]
time = 0.00, size = -1, normalized size = -0.01 \begin {gather*} \int \frac {x^2}{{\left (a-b\,x^4\right )}^{5/4}} \,d x \end {gather*}

Verification of antiderivative is not currently implemented for this CAS.

[In]

int(x^2/(a - b*x^4)^(5/4),x)

[Out]

int(x^2/(a - b*x^4)^(5/4), x)

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